In fact, you just need the derivative w.r.t. $m$ of $(x/a)^m I_m(ax)$, right ?
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LierreNov 9 '12 at 15:44

Thanks Lierre, you are right, I was just not sure about that because it is a derivative under the integral sign...but you are right... it the derivative of $(x/a)^m I_m(ax)$ w.r.t $m$
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RemyNov 9 '12 at 17:17

1 Answer
1

For large $x$, if $a>0$, $I_m$ behaves asymptotically like $I_m(ax)\approx e^{ax}/\sqrt{ax}$. Therefore for large $x$ the integrand will look like $x^{m-1/2}e^{-(x-a)^2/2}$. This dies off fast enough that the improper integral converges uniformly in $m$ and you can differentiate inside the integral.